Homework Help: Calculating Moments, Equilibrium

1. Nov 27, 2007

jmcmillian

Hi guys,

I am having difficulty understanding moments and equilibrium.

I have attached a diagram of a sample problem. The red Arrows are T1 resolved into X and Y components.

Let's say that I want to find the equilibrium equation for the moments about A and B. I realize that forces going through the point (although there is none in this particular diagram) are zero at that point. I also realize that this process is going to involve multiplying the force times it's distance relative to the point of which the moment is about.

So in other words, finding the equilibrium equation of the moments about A, it would be something like

Moments about A: (600N)(2M) (T1sin(theta))(5m) (100N)(3.25M)

But what I struggle with is determining which are negative and which are positive. I really have struggled grasping the right hand rule, and that's likely my problem.

Any help/advice/criticism would be greatly appreciated.

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2. Nov 27, 2007

rl.bhat

You can take clockwise moments as positive and anticlockwise moments as negative.

3. Nov 27, 2007

jmcmillian

thanks for your help. I realize that, but how do I determine which moments are clockwise and which are anticlockwise?

4. Nov 27, 2007

rl.bhat

Direction in which the needle of the clock moves is clockwise motion.

5. Nov 27, 2007

jmcmillian

so in the case of the diagram given, 600 N would be clockwise from A, T1sin(theta) would be anticlockwise?

6. Nov 27, 2007

Astronuc

Staff Emeritus
As rl.bhat indicated, 'clockwise' is the direction in which second, minute and hour hands move on a clock/watch.

Or, it assumes a right-hand rule whereby one looks into the page/plane and rotates the top of the hand to the right, which would be the direction one would turn a screw (righthanded thread) to tighten it. Alternatively, counterclockwise would losen a screw or back it out of the hole.

7. Nov 27, 2007

jmcmillian

Thanks, Astronuc. How is that applied to forces that are simply given horizontal or vertical directions? To determine the moment, is it from the force to the point of the moment (I.e. like from 600N to point A, which would be counterclockwise? Or, are we looking from the point to the force?

8. Nov 27, 2007

Astronuc

Staff Emeritus
A force acting along (parallel to the beams axis) will provide a tensile or compressive force, depending one whether the forces stretches (pulls) or compresses (pushes) the beam.

A moment relates to the force component which is perpendicular to beam axis at some moment arm (distance from the pivot).

See if this helps - http://www.ecf.toronto.edu/apsc/courses/civ214/Lectures/Sign%20convention%20for%20moment%20diagrams.html [Broken]

With respect to a pivot at A, the 600 N is clockwise, T1y is counterclockwise.

Is one trying to develop a moment diagram?

Last edited by a moderator: May 3, 2017
9. Nov 27, 2007

jmcmillian

Just Checking for Understanding:

resolving each force into x and y components would be my first order of business. Then, finding the perpendicular distance between the point of which I am pivoting.

Thanks for the Toronto Site. it has been helpful. I do have a question about a particular statement of it, and I think this is the question I really have been shooting for:

"It is more straightforward to define that moments are to be drawn on the tension side of members for all components of a given structure"

So....I need to determine clockwise or counterclockwise based on what it would take to PULL the force to the pivot?

Let's say I have a point C on a bar and a force on that same bar 6 meters from the right of it, in a vertically down direction (pulling the bar). Declaring counterclockwise as positive, My moment arm would be perpendicular in the clockwise (negative) direction, because it is shorter (90 degrees as opposed to 270)?

10. Nov 27, 2007

Astronuc

Staff Emeritus
No, that's not what the statment means.

If one has beam, with two supports (pivots) at each end, and a force pushing down in the middle (i.e. between the two supports), then the top of the beam will be in compression and the bottom of the beam will be in tension.

Let me see if I can find some other examples.

11. Nov 28, 2007

jmcmillian

Sorry Astronuc, I did not see this last part until just now. Does this mean that if the force is down and to the right of the pivot point, it is clockwise, whereas if it is up and to the right, it is counterclockwise?

As to my objective in all of this, I am trying to find equilibrium equations for the forces at each moment. The image I attached is a sample that I made up, but it is somewhat similar to my assignment.

12. Nov 28, 2007

Astronuc

Staff Emeritus
Yes.

Perhaps these will help.

http://em-ntserver.unl.edu/NEGAHBAN/Em325/11-Bending/Bending.htm [Broken]

http://em-ntserver.unl.edu/NEGAHBAN/Em325/10a-shear-and-bending-moment/Shear%20stress%20in%20beams.htm [Broken]

In general - http://em-ntserver.unl.edu/NEGAHBAN/Em325/intro.html [Broken]

Last edited by a moderator: May 3, 2017
13. Nov 29, 2007

jmcmillian

Thank you again for all of your efforts. I really think that it is starting to make sense now.

PF has been my backbone this year - my instructor hasn't been very available for help, and gets impatient when you don't understand what he's saying. Having the class at 8am doesn't allow for maximum concept registry, either :)

Last edited by a moderator: May 3, 2017